Double Factorials and multifactorials

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Before reviewing this article, the readers are expected to know what factorials are first.

The double factorial of a positive integer \(n\) is the generalization of the factorial \(n!;\) this type of factorial is denoted by \(n!!\). It is a type of multifactorial which will be discussed later. As far as double factorial is concerned, it ends with \(2\) for an even number, and ends with \(1\) for an odd number:

For an even number and \(n>0\), \(n!!=n\times (n-2)\times \cdots\times 4\times 2.\)

For an odd number and \(n>0\), \(n!!=n\times (n-2)\times \cdots\times 3\times 1.\)

If \(n=0\), \(0!!=1.\)

To clarify, \(n!! \) is not equal to \((n!)! \). For example, \(4!! = 4\times2=8\), whereas \((4!)! = 24! = 620448401733239439360000\). Go ahead and try out the following warm-up problem:

For any non-negative integer \(n,\) we find that:\[\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!. \ _\square\]

Here, there is no need to consider two separate cases because it makes no difference whether \(n\) is odd or even.

We can expand the LHS as \[\dfrac{(2n+1)\times (2n)\times (2n-1)\times \cdots \times 3\times 2\times 1}{(2n)\times (2n-2)\times (2n-4)\times \cdots \times 4\times 2}.\] Since all the even numbers \(2n, 2n-2, 2n-4, \ldots, 4, 2\) get canceled, we are left with the equation \[\dfrac{(2n+1)!}{(2n)!!}=(2n+1)!!. \ _\square\]

Try the first part here!

For any non-negative integer \(n\) we find that

\[\dfrac{(2n-1)!}{(2n-2)!!}=(2n-1)!!. \ _\square\]

Again here, there is no need to consider two separate cases. We can expand the LHS as \[\dfrac{(2n-1)\times (2n-2)\times (2n-3)\times \cdots \times 3\times 2\times 1}{(2n-2)\times (2n-4)\times \cdots \times 4\times 2}.\] Since all the even numbers \(2n-2, 2n-4, \ldots , 4, 2\) get canceled, we are left with the equation \[\dfrac{(2n-1)!}{(2n-2)!!}=(2n-1)!!. \ _\square\]